Question

Attempt either (a) or (b): **(a)** Calculate the total resistance of the given circuit and find the total current flowing in it. (Circuit: $R_1 = 1\,\Omega$ in series with parallel combination of $R_2 = 10\,\Omega$ and series branch $R_3 + R_4 = 6\,\Omega + 4\,\Omega = 10\,\Omega$, connected to a $12	ext{ V}$ battery.) **OR** **(b)** How would you join three resistors each of resistance $6\,\Omega$, so that the equivalent resistance of the combination is: (i) $9\,\Omega$ (ii) $4\,\Omega$?

Accepted Answer

2. **Part (a):** - $R_3$ and $R_4$ are in series: $R_{34} = 6 + 4 = 10\,\Omega$. - $R_{34}$ is in parallel with $R_2 = 10\,\Omega$: $$R_p = \frac{10 	imes 10}{10 + 10} = 5\,\Omega$$ - Total resistance $R_{	ext{total}} = R_1 + R_p = 1 + 5 = 6\,\Omega$. - Total current $I = \frac{V}{R_{	ext{total}}} = \frac{12}{6} = 2	ext{ A}$. **Part (b):** (i) $9\,\Omega$: Two resistors in parallel connected in series with the third ($6 \parallel 6 + 6 = 3 + 6 = 9\,\Omega$). (ii) $4\,\Omega$: Two resistors in series connected in parallel with the third ($(6 + 6) \parallel 6 = 12 \parallel 6 = 4\,\Omega$).

Circuit Total Resistance and Resistor Combinations

Explanation

Practise this under board exam conditions